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Gwar [14]
3 years ago
7

In parallelogram ABCD shown, point E and F are located on diagonal BD and point G is located on side AB such that GE and CF are

perpendicular to BD. Prove: trainge BEG is similar to traingle DFC​

Mathematics
1 answer:
Lena [83]3 years ago
8 0

Answer:

Match the Statement number with the Reason number.

Step-by-step explanation:

Statements:

1.GE is perpendicular to BD

CF is perpendicular to BD

2. Angle GED equal 90

Angle DFC equal 90

3. Angle GED is congruent to Angle DFC

4. Angle GBE is congruent to Angle FDC

5. Triangle BEG is similar to DFC

Reasons:

1.Given

2. Definition of perpendicular lines

3. Transitive property

4. Alternate Interior Angles

5. AA Similarity

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Solve the equation. dy dx = ay + b cy + d , where a, b, c, and d are constants. (assume a ≠ 0 and ay + b ≠ 0.)
Reil [10]

 

It solves it in x. the solution for y includes heavy use of the product log function.

dy/dx                                = ay + b/cy +d

(cy +d/ ay + b) dy            = dx

∫ (cy +d/ ay + b) dy          = x (t) + C

 

Into solving the integral, integration by parts followed by u substitution and another integration by parts.

 

∫ (cy +d/ ay + b) dy

u            = cy + d dv          = dy/ay + b

du          = c dy v               = ln I ay + b l / a

 

Then, use u substitution for the new integral

 

u            = ay + b

du          = a dy

∫ ln l ay + b I dy                = ∫ ln IuI /a du    = 1/a ∫ ln luI du

 

Integrating the natural log includes thus far another integration by parts

r             = ln IuI ds            = du

dr          = du / u (s)           = du

∫ ln IuI / du                         = u ln IuI - ∫ du   = u ln IuI - ∫ a dy                                                                                   = (ay + b) ln Iay +bl – ay

 

The original integral of expression

∫ (cy +d/ ay + b) dy             = cy + d/a ln lay+bl – c/a² [(ay+b) ln lay+bl – ay]

Then simplify

∫ (cy +d/ ay + b) dy             = cy + d/a ln lay+bl – c/a²[(ay+b) ln lay+bl – ay]

                                           = a (cy + d)/a² ln lay+bl – c (ay+b)/ a²ln lay+bl +                                                               c/a² ay

                                           = cay + ad – cay – cb/ a² ln lay+bl + cay/a²

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 The final answer will be

x(t) + C                               = ad – cb/a² ln lay+bl + cy/a

x(t)                                     = ad – cb/a² ln lay+bl + cy/a + k

 

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